Question
Factorise : $(a^2- a) (4a^2- 4a - 5) - 6$
✓
Answer
Let us assume,
$a^2 - a = x$
Then the given expression is,
$(a^2- a) (4a^2- 4a - 5) - 6$
$= x( 4x - 5 ) - 6$
$= 4x^2 - 5x - 6$
$= 4x^2 - 8x + 3x - 6$
$= 4x( x - 2 ) + 3( x - 2 )$
$= ( 4x + 3 )( x - 2 )$
$= [ 4( a^2 - a ) + 3 ]( a^2 - a - 2 ) [$ resubstitute the value of $x ]$
$= [ 4a^2 - 4a + 3 ]( a^2 - a - 2 )$
$=[ 4a^2 - 4a + 3 ]( a^2 - 2a + a - 2 )$
$= [ 4a^2 - 4a + 3 ][ a( a - 2 ) + 1( a - 2 )]$
$= [ 4a^2 - 4a + 3 ]( a - 2 )( a + 1 )$
$a^2 - a = x$
Then the given expression is,
$(a^2- a) (4a^2- 4a - 5) - 6$
$= x( 4x - 5 ) - 6$
$= 4x^2 - 5x - 6$
$= 4x^2 - 8x + 3x - 6$
$= 4x( x - 2 ) + 3( x - 2 )$
$= ( 4x + 3 )( x - 2 )$
$= [ 4( a^2 - a ) + 3 ]( a^2 - a - 2 ) [$ resubstitute the value of $x ]$
$= [ 4a^2 - 4a + 3 ]( a^2 - a - 2 )$
$=[ 4a^2 - 4a + 3 ]( a^2 - 2a + a - 2 )$
$= [ 4a^2 - 4a + 3 ][ a( a - 2 ) + 1( a - 2 )]$
$= [ 4a^2 - 4a + 3 ]( a - 2 )( a + 1 )$
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